Math 711: Lecture of October 19, 2005
To prove that the Plücker relations hold, we think of the columns indexed by i1 , . . . , ia
and k1 , . . . , kb as fixed. The left hand side may then be thought of as a function of m
vector variables in K r , namely the columns indexed by j1 , . . . , jm . We first show that if
we let π range over all permutations of 1, . . . , m (writing πc for π(c)), we have that
(#)
X
sgn (π)X[i1 , . . . , ia , jπ1 , . . . , jπt ]X[jπt+1 , . . . , jπs , k1 , . . . , kb ] = 0.
π∈Sm
To prove this we observe that the left hand side is multilinear in the m variable columns
indexed by j1 , . . . , jm . It is also alternating: if two columns are equal, any summand in
which they both occur in the first factor or in the second factor is 0, while for each term
where one occurs in the first factor and one in the second factor, there is a corresponding term, which results from switching the relevant indices,
has the opposite sign.
Vm which
Therefore, the left hand side represents a map from
(K r ) to K, and is therefore 0,
because m > r. We shall complete the proof that the Plücker relation holds by showing
that the left hand side of (#) is t!u! times the left hand side of the Plücker relation. The
point is that for each t element subset T consisting of jν1 < · · · < jνt in {1, . . . , m}, there
is a unique summand in the Plücker relation in which that subset occurs, but there are t!u!
permutations π such that the subscripts with those columns occur on the left. To bring the
term involving π to the term involving ν, we need to apply the permutation ν ◦ π −1 . Note
that π, ν and, hence, ν ◦ π −1 all permute T and permute its complement T 0 in {1, . . . , m}.
The sign of ν ◦ π −1 is the product of the signs of the separate permutations it induces on
T and T 0 . Thus, all t!u! terms can be written as sgn (π)sgn (ν ◦ π −1 ) = sgn (ν) times the
product of the two minors in the term corresponding to ν, and so all of these terms are
the same as the term in the Plücker relation. Thus, when we multiply the left hand side
of the Plücker relation by t!u!, we get 0. Since we are working in equal characteristic 0, we
have proved what we need. We shall next prove that the Plücker relations can be used to write any product hh0 of
incomparable minors as a linear combination of standard quadratic monomials in which,
for every term, the smaller of the two minors occurring is < h. This will suffice, because
the same reasoning applied to h0 h then shows that it can be written in the same way with
the smaller minor in each term ≤ h0 . By the linear independence of standard monomials,
these two reperesentations are actually the same. Therefore, the smaller minor in each
term is less than both h and h0 . (The argument actually shows as well that the larger
minor in each terms is greater than both h and h0 .)
Before giving the argument we note the following fact: if i1 < · · · < ir , and one or more
of these integers is replaced by a smaller integer (distinct from the remaining it ), the new
minor (with its columns arranged with indices in increasing order), is strictly less than
X[i1 , . . . , ir ]. It suffices to see this for just one replacement. Suppose that in is replaced
by i, where ic < i < ic+1 ≤ in . The ij do not change for j ≤ c or j > n, but ic+1 is
1
2
replaced by i, decreasing, and ij is replaced by ij−1 for c + 1 < j ≤ n. Entirely similarly,
if one or more of the ij is replaced by a larger integer (distinct from the remaining it ), the
new minor (with its columns arranged with indices in increasing order) is larger than h.
We are now ready to carry through the proof that Plücker relations suffice for straightening. Suppose that h = X[i1 , . . . , ir ] and h0 = [k1 , . . . , kr ] give a counter-example, where
i1 < · · · < ir , k1 < · · · < kr and for some a < r we have that in ≤ bn for 1 ≤ n ≤ a while
ia+1 > ka+1 (a may be 0). Choose the counter-example with a as large as possible. Take
m = r + 1, and apply the Plücker relation for
X[i1 , . . . , ia , ia+1 , . . . , ir ]X[k1 , . . . , ka+1 , ka+2 , . . . , kr ],
where the sequence ka+2 , . . . , kr is empty if a = r − 1. Here, the elements j1 , . . . , jm are
ia+1 , . . . , ir , k1 , . . . , ka+1 . The terms in the Plücker relation other than the one displayed
above all have some of the elements ia+1 , . . . , ir replaced by a corresponding number of
elements from k1 , . . . , ka+1 , all of which are ≤ ia+1 , and all of which are therefore smaller
than the elements they are replacing. It follows that the smaller minor in each term is
< h. Similarly, the larger minor is > h0 . Suppose that a typical summand is, up to sign,
X[f1 , . . . , fr ]X[g1 , . . . , gr ].
Since X[f1 , . . . , fr ] < h and h0 < X[g1 , . . . , gr ], we certainly have fi ≤ gi for i ≤ a.
But it is also the case that (after arranging the indices for each factor so that they are
increasing) one has that fa+1 ≤ ga+1 . To see why, note that the elements f1 , . . . , fr
include at least a + 1 terms tthat are ≤ max {ia , ka+1 } so that fa+1 ≤ max {ia , ka+1 },
while the elements g1 , . . . , gr include at least r − a terms that are ≥ min {ia+1 , ka+2 }, so
that ga+1 ≥ min {ia+1 , ka+2 } > max {ia , ka+1 } ≥ fa+1 , as required.
By induction on a we can reach the required form. A poset H is called bounded if it has a greatest and least element. A poset H is called
a distributive lattice if it is bounded, if any two elements x, y have a least upper bound
x ∨ y and and a greatest lower bound x ∧ y, and the following axioms are satisfied:
(1) For all x, y, z ∈ H, x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z).
(2) For all x, y, z ∈ H, x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z).
Either axiom implies the other. E.g., assume (1). Then with w = x ∨ y, we have that
(x ∨ y) ∧ (x ∨ z) = w ∧ (x ∨ y) = (w∧ x) ∨ (w ∧ z) = x ∨ (w ∧ z) = x ∨ (x ∨ y) ∧ z =
x ∨ (x ∧ z) ∨ (y ∧ z) = x ∨ (x ∧ z) ∨ (y ∧ z) = x ∨ (y ∧ z), as required.
We say that y is a cover of x in a poset H if y > x and no element is strictly between
y and x.
A poset is called locally upper semimodular if whenever y and z are covers of x and there
exists v such that x < v and y < v, then there exists w ≤ v such that w is a cover of both
x and y.
Note that the poset of r ×r minors of the r ×s matrix H is a distributive lattice. In fact,
if h = X[i1 , . . . , ir ] and h0 = X[j1 , . . . , jr ], where i1 < · · · < ir and j1 < · · · < jr , it is not
diificult to see that h ∨ h0 = X[i1 ∨ j1 , . . . , ir ∨ jr ] and h ∧ h0 = X[i1 ∧ j1 , . . . , ir ∧ jr ].
The only point of interest is that one needs to see that the sequences produced in this
3
way are still strictly increasing, which is straightforward. Note that H has least element
X[1, . . . , r] and greatest element X[s − r + 1, . . . , s].
We shall prove that a distributive lattice is locally upper semimodular and that an ALS
over a field K such that the poset is bounded and locally upper semimodular is CohenMacaulay. Along with the comments in the preceding paragraph, this will establish that
K[X/r] is Cohen-Macaulay when K is any field and, hence, when K is any Cohen-Macaulay
ring. Before doing so, we analyze the dimension of K[X/r].
The rank or height of an element of a poset is the supremum of lengths of chains
descending from that element. We claim that the height of X[i1 , . . . , ir ] is
r
X
t=1
it −
r
X
t=1
r=
r
X
t=1
it −
r(r + 1)
.
2
Any maximal chain descending from X[i1 , . . . , ir ] must end with X[1, . . . , r] The assetion follows if we can show that given consecutive elements in the chain, the sum of the
entries decreases by exactly one from the larger to the smaller. Suppose X[j1 , . . . , jr ] >
X[k1 , . . . , kr ] are consecutive. Consider the smallest t such that kt < jt . Then
X[j1 , . . . , jt−1 , jt − 1, jt+1 , . . . , jr ]
will be strictly between X[j1 , . . . , jr ] and X[k1 , . . . , kr ] unless it is equal to X[k1 , . . . , kr ]
(note that jt−1 = kt−1 < kt ≤ jt − 1 here).
It follows
Pr that the length of a longest chain in H is the height of X[s − r + 1, . . . , s],
which is t=1 (s − r + t − t) = r(s − r), and this is the dimension of the order complex.
Hence:
Corollary. For any ring K, if X is an r × s matrix of indeterminates over K with
1 ≤ r ≤ s, dim (K[X/r]) = dim (K) + r(s − r) + 1.